Fluid Mechanics
Chapter 9
Defining a Fluid
• A fluid is a nonsolid state of matter in which
the atoms or molecules are free to move past
each other, as in a gas or a liquid.
• Both liquids and gases are considered fluids
because they can flow and change shape.
9-1: Fluids and Buoyant Force
• Liquids have a definite volume
• Gasses do not
• Liquids do not have a definite shape
Density and Buoyant Force
• In general, density is a measure of how
much there is of a quantity in a given
amount of space.
– Quantity can be anything; # of trees or
people, or amount of mass or energy
Mass Density
• The concentration of matter of an object is
called the mass density.
(Amount of mass per unit volume of a substance)
• Represented with the Greek symbol, ρ (rho)
Mass Density = mass/volume
ρ = m/v
S.I. Unit Kg / m3
Common fluids
and solids
• Solids and liquids tend
to be almost
incompressible.
i.e., change very
little with changes
in pressure.
Buoyant Force
• Buoyant Force – a force that acts upward
on an object submerged in a liquid or
floating on a liquid’s surface.
• This force can keep objects afloat
• Acts in direction OPPOSITE of gravity
Buoyant Force
Archimedes Principle
• Law of Archimedes:
The buoyant force is equal to the
weight of the replaced liquid or gas.
Buoyant Force and Archimedes’
Principle
• The Brick, when added will cause the water to be
displaced and fill the smaller container.
What will the volume be inside the smaller
Container?
The same volume as the brick!
Buoyant Force and
Archimedes’ Principle
Archimedes’ principle describes the magnitude
of a buoyant force.
Archimedes’ principle: Any object completely or
partially submerged in a fluid experiences an
upward buoyant force equal in magnitude to the
weight of the fluid displaced by the object.
FB = Fg (displaced fluid) = mfg
magnitude of buoyant force = weight of fluid displaced
Buoyant Force
The raft and cargo
are floating
because their
weight and
buoyant force are
balanced.
Buoyant Force
• Now imagine a small hole
is put in the raft.
• The raft and cargo sink
because their density is
greater than the density of
the water.
• As the volume of the raft
decreases, the volume of
the water displaced by the
raft and cargo also decreases,
as does the magnitude of the
Buoyant force.
Buoyant Force
• For a floating object, the buoyant force equals the object’s
weight.
•
The apparent weight of a submerged object depends on the
density of the object.
•
For an object with density ρo submerged in a fluid of density
ρf, the buoyant force FB obeys the following ratio:
Fg (object) = ρo
FB
ρf
Example
A bargain hunter purchases
a “gold” crown at a flea
market. After she gets
home, she hangs the crown
from a scale and finds its
weight to be 7.84 N. She
then weighs the crown while
it is immersed in water, and
the scale reads 6.86 N. Is
the crown made of pure
gold? Explain.
Solution
Choose your equations:
Fg – FB = apparent weight
Fg = ρo
FB
ρf
Rearrange your equations:
FB = Fg – (apparent weight)
Ρo = Fg (ρf)
FB
Plug and Chug
FB = 7.84 N – 6.86 N = 0.98 N
Ρo =
7.84 N (1.00 10 kg/m) = 8.0 x 103 kg/m3
0.98 N
• From the table in your book, the density of
gold is 19.3 x 103 kg/m3.
• Because 8.0 x 103 kg/m3 < 19.3 x 103 kg/m3,
the crown cannot be pure gold.
HOMEWORK
• Page 324
– Practice 9A # 1 and 3
– Section Review # 1-3
QOTD: Wed 1/13/2010
Get a calculator from the red bin and complete
the following:
1: Calculate the actual weight, buoyant
force, and apparent weight of a 5.0 X 10-5
m3 iron ball floating at rest in mercury.
(density of mercury is 13.6 x 103 kg/m3)
2: How much of the ball’s volume is
immersed in mercury?
QOTD: Wed 1/13/2010
• Calculate the actual weight, buoyant force,
and apparent weight of a 5.0 X 10-5 m3 iron
ball floating at rest in mercury. (density of
mercury is 13.6 x 103 kg/m3)
3.86N, 3.86 N, 0 N
• How much of the ball’s volume is
immersed in mercury?
2.89 x 10-5 m3
Fluid Pressure and Temperature
• Objectives:
– Calculate pressure exerted by a fluid
– Calculate how the pressure varies with depth
in a fluid
– Describe fluids in terms of temperature
PRESSURE
• THE MAGNITUDE OF THE FORCE ON A
SURFACE PER UNIT AREA
• PRESSURE = FORCE/AREA
P=F
A
PRESSURE
SI unit is the pascal (Pa)
1 pascal is equal to 1 N/m2
The pressure of the atmosphere at sea level is
about 105 Pa
* this amount of air pressure under
normal conditions is the basis for
another unit, the atmosphere (atm)
The absolute air pressure inside a typical
automobile tire is about 300,000 Pa, or about 3
atm.
Applied Pressure
• Transmitted equally throughout a fluid
• In general, if the pressure in a fluid is
increased at any point in a container, the
pressure increases at all points inside the
container by exactly the same amount.
• This is noted by PASCAL
• Which leads us to PASCAL’S PRINCIPLE
PASCAL’S PRINCIPLE
• PRESSURE APPLIED TO A FLUID IN A
CLOSED CONTAINER IS TRANSMITTED
EQUALLY TO EVERY POINT OF THE
FLUID AND TO THE WALLS OF THE
CONTAINER.
Pascal’s Principal
• A hydraulic lift uses this principle.
• A small force, F1, applied to a small piston of
area A1 causes a pressure increase in a fluid,
such as oil.
• This increase in pressure, Pinc, is transmitted to
a larger piston of area A2 and the fluid exerts a
force, F2, on this piston.
See figure 9-7 in text
• Applying Pascal’s principle and the
definition of pressure gives us the
following equation :
Pinc = F1 = F2
A1
A2
Rearranging the equation to solve for F2:
F2 = F1 A2
A1
From this equation, we see
that F2 is greater than F1
by a factor equal to the
ratio of the areas of the 2
pistons.
Pascal’s Principle
Because the pressure is the same on both sides of
the enclosed fluid in a hydraulic lift, a small force
on the smaller piston (left) produces a much larger
force on the larger piston (right).
Pressure
Pressure varies with depth in a fluid
• Water pressure increases with depth
because the water at a given depth must
support the weight of the water above it.
Fluid pressure as a
function of depth
• By using the symbol ρo for the atmospheric
pressure at the surface, we can express the
total pressure, or absolute pressure, at a given
depth in a fluid of uniform density, ρ, as follows:
ρ = ρo + ρgh
Absolute pressure = atmospheric pressure + (density x free fall accel x depth)
Temperature
• The temperature in a gas can be understood as
what is happening to the atomic scale.
• The higher the temperature of a gas, the faster
the particles move.
• The faster the particles move, the higher the rate
of collisions against the walls of the container.
• More momentum is transferred to container
walls in given time interval, thus an increase in
pressure results.
SI units for Temperature:
Kelvins and degrees Celsius
To convert from Celsius to Kelvin, add 273.
Ex: Room temp is about 20° Celsius or 293 k.
HOMEWORK
Pg 327 # 1-2
Pg 330 # 1
Pg 331 # 1, 3-4
QOTD Friday: 1/15/2010
FLUIDS IN MOTION
• FLUID FLOW:
– When a fluid is in motion, the flow can be
characterized in one of two ways. The flow is
said to be laminar if every particle that passes a
particular point moves along the same smooth
path traveled by the particles that passed that
point earlier.
– This path is called streamline.
– Different streamlines CANNOT cross each other.
Fluid Flow Cont.
Flow of fluid becomes irregular, or turbulent,
above a certain velocity or under conditions
that can cause abrupt changes in velocity.
– obstacles or sharp turns
• Irregular motion of the fluid – eddy currents
– Turbulent flow
• Ex: water in wake of ship or in air currents of severe
thunderstorm
• Ideal Fluid : fluid that has no internal
friction or viscosity and is incompressible
• Viscosity: amount of internal friction within
a fluid.
– Can occur when one layer of fluid slides past another.
Principles of Fluid Flow
• The mass flowing into the pipe MUST equal
the mass flowing out of the pipe in the same
time interval. Thus, m1 = m2
• Remember…m = pv
• Volume of a cylinder: V = A (delta X)
So,
p1V1 = p2V2
p1A1 (Delta X 1) = p2A2 (Delta X 2)
p1V1 = p2V2
p1A1 (Delta X 1) = p2A2 (Delta X 2)
*The length of the cylinder is also the distance the
fluid travels, which is equal to the speed of flow
multiplied by the time interval (delta X = v delta t)
• So,
p1A1 v1 (Delta t) = p2A2 v2 (Delta t)
** for an ideal fluid, both the time interval and
the density are the same on each side of the
equation, so they cancel each other out.
p1A1 v1 (Delta t) = p2A2 v2 (Delta t)
Continuity Equation
• After canceling each other out, we are left with
the CONTINUITY EQUATION:
A1v1 = A2v2
Simply state:
area x speed in region 1 = area x speed in region 2
• Flow rate is CONSTANT throughout a pipe
• Pressure in a fluid is related to the speed
of flow.
– BERNOULLI’S PRINCIPAL:
The pressure in a fluid decreases
as the fluid’s velocity increases
Bernoulli’s Equation cont.
• Explains the conservation of energy in
fluids. Expressed as:
P + ½ ρv2 + ρgh = constant
Pressure +KE per unit volume + PEg per unit volume =
constant along a given streamline
Homework
• Pg 337: Practice #1
• Section Review # 1
QOTD Tuesday 1/19/2010
• What is the continuity equation?
• Using the equation, if a pipe narrows from a
cross section of 2.0 m2 to 0.30 m2 and the
speed through the wider area is 8.0 m/s,
what is the speed of the water flowing
through the narrow part?
9-4: Properties of Gases
• Volume, Pressure, and Temperature are
the three variables that completely
describe the macroscopic state of an ideal
gas.
IDEAL GAS LAW
• PV = NkBT
Pressure x volume = # gas particles x Boltzmann’s
constant x temperature
KB = 1.38 x 10-23 J/K
Temperature MUST be in Kelvins!!!!
IDEAL GAS LAW
• PV = nRT
Pressure x volume = # moles of gas x molar
gas constant x temperature
n = # moles of gas
Where one mole = 6.02 x 1023 particles
R = 8.31 J/(mol * K)
IDEAL GAS LAW
• If the # of gas particles remains constant,
then…
P1V1 = P2V2
T1
T2
Sample
• A sealed tank with a volume of 0.10 m3
contains air at 27 degrees Celsisus under
pressure 18000 Pa. The valve can
withstand pressures up to 50000. Will the
valve hold if the air inside the tank is
heated to 227 degrees C?
Sample
• A sealed tank with a volume of 0.10 m3 contains air at 27
degrees Celsius under pressure 18000 Pa. The valve
can withstand pressures up to 50000. Will the valve hold
if the air inside the tank is heated to 227 degrees
Celsius?
P 1V 1 = P 2V 2
T1
T2
Cont…
18000 (0.10) = (P2) (0.10)
300
500
900,000 = 30 (P2)
P2 = 30,000 Pa
Thus, yes the valve will hold because
30,000 Pa is LESS THAN 50,000 Pa max
that it can withstand.
Homework
• Page 341: # 1-2
• Page 341 Review: #1 and 5
• STUDY FOR YOU EXAM!!!
• Make sure you know the summary and
Key Ideas found on page 342 in your text.